Question 14.12: Calculate the molarity of a sodium hydroxide solution that i...

Calculate the molarity of a sodium hydroxide solution that is prepared by mixing 100. mL of 0.20 M NaOH with 150. mL of water. Assume that the volumes are additive.

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Step 1:
Calculate the moles of NaOH in the original solution. We are given that the volume of NaOH is 100 mL and the molarity is 0.20 M. We can use the formula moles = volume x molarity to find the moles of NaOH. In this case, the volume is 100 mL (or 0.100 L) and the molarity is 0.20 M. So, moles = (0.100 L) x (0.20 mol NaOH/1 L) = 0.020 mol NaOH.
Step 2:
Divide the moles of NaOH by the final volume of the solution to find the new molarity. The final volume of the solution is given as 250 mL (or 0.250 L). So, molarity = moles/volume = 0.020 mol NaOH / 0.250 L = 0.080 M NaOH.
Step 3:
Check the answer. We know that if we double the volume of the solution, the concentration is halved. In this case, the original volume is 100 mL and the final volume is 250 mL, which is more than double. Therefore, the concentration of the new solution should be less than 0.10 M. Since our calculated molarity is 0.080 M, it is indeed less than 0.10 M, confirming our answer.

Final Answer

READ    Knowns  V_2 = 150. mL water

V_1 = 100. mL NaOH

V_{solution} = 250. mL

M_{initial} = 0.20 M

PLAN              In the dilution, the moles NaOH remain the same; the molarity and volume change.

1. Calculate the moles of NaOH in the original solution.

mol = (0.100    \cancel{L}) (\frac{0.20   mol   NaOH}{1  \cancel{L}} )= 0.020 mol NaOH

2. Divide the mol NaOH by the final volume of the solution to find the new molarity.

CALCULATE:  M= \frac{0.020   mol   NaOH}{0.250   L} =0.080 M NaOH

CHECK      If we double the volume of the solution, the concentration is half. Therefore the concentration of the new solution here should be less than 0.10 M.

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